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Given a wire and a piece of foil wired to a bread board, is it possible using capacitors wired in series (or some other method) to significantly reduce the capacitance of the wire/foil?

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  • \$\begingroup\$ Do you need to reduce the DC capacitance or the AC capacitance? \$\endgroup\$ – Ignacio Vazquez-Abrams Aug 8 '13 at 2:04
  • \$\begingroup\$ :( I don't know. I have a capacitance sensor, I'm trying to reduce the value on it. \$\endgroup\$ – j03m Aug 8 '13 at 2:14
  • \$\begingroup\$ I believe DC... \$\endgroup\$ – j03m Aug 8 '13 at 2:16
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    \$\begingroup\$ A photo of what you have now might help us understand your problem. \$\endgroup\$ – The Photon Aug 8 '13 at 3:22
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    \$\begingroup\$ If I've got the right mental picture of what you're doing, you have two choices: reduce the size of the foil, and move all other conductors further away from your wire and foil. \$\endgroup\$ – The Photon Aug 8 '13 at 3:23
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I'm not 100% sure I understand your needs, but you might consider a negative capacitance amplifier, as in http://en.wikipedia.org/wiki/Negative_impedance_converter, whose parallel negative input impedance should subtract that of your wire. This is a positive feedback trick, so if your negative capacitance is too high, your system will go unstable.

This is used all the time in neuroscience to deal with large capacitances in glass microelectrodes, but I've never seen it used for cable capacitance.

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I've used double screened cable. The inner core signal is applied to a unity gain amplifier which projects back the same voltage onto the 1st (inner screen). This means that the inner core "sees" a capacitance that is vastly reduced. The outer screen is used as ground and of course the inner screen "sees" all the ground capacitance.

I've used this on a capacitance probe too.

Things that make it go wrong - long wires where the frequency used has a wavelength that approaches (maybe) 100th of the cable length. Significant attenuation down the cable (losses) means it isn't as effective.

Use high impedance inputs on your actual cap-probe input and the non-inverting screen driver input.

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  • \$\begingroup\$ Not quite "doesn't see any capacitance", but rather "reduces current caused by the cable capacitance". If i=CdV/dt, then one can reduce the current by reducing C, or by reducing dV/dt. Double shielded cable with an amp driving the inner shield at the common mode voltage achieves the latter. \$\endgroup\$ – Scott Seidman Oct 7 '13 at 12:34
  • \$\begingroup\$ I think you meant to say " Inner screen" here " ...the inner core sees all the ground capacitance." \$\endgroup\$ – placeholder Oct 7 '13 at 12:52

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